1.For the function f(x)={1x−2+(x+2)(12)x−2,x≠2k,x=2f(x) = \begin{cases} \dfrac{1}{x-2} + (x+2)\left(\dfrac{1}{2}\right)^{x-2}, & x \neq 2 \\[8pt] k, & x = 2 \end{cases}f(x)=⎩⎨⎧x−21+(x+2)(21)x−2,k,x=2x=2, which of the following holds?a.k=12k = \dfrac{1}{2}k=21 and fff is continuous at x=2x = 2x=2b.k≠0,12k \neq 0, \dfrac{1}{2}k=0,21 and fff is continuous at x=2x = 2x=2c.fff cannot be continuous at x=2x = 2x=2d.k=0k = 0k=0 and fff is continuous at x=2x = 2x=2Login to continueOnly logged in users canattempt or see the solution.
